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Inserted text in green WW WM HEADERS_END The distance to the moon can be computed as follows. We know that: * acceleration due to gravity at the Earth's surface is about EQN:9.8ms^{-2}. ** this can be measured using a pendulum in your living rioom ** this can be measured using a pendulum in your living room * the orbital period of the moon relative to the stars is about 27.3 days. * acceleration due to gravity decreases as the inverse square. * the radius of the Earth is 6366km. ** Measured by Eratosthenes OK, so acceleration in a circle is EQN:v^2/r or EQN:\omega^2r where /r/ is the radius of the orbit, and EQN:\omega is the rotational velocity. That's 27.3 days divided by EQN:2\pi giving 1/375402 radians/second. Hence acceleration in the orbit is EQN:(1/375402)^2r where /r/ is the (unknown) distance from the centre of the Earth to the Moon. But acceleration due to gravity is EQN:9.8(r/R)^{-2} where /R/ is the radius of the Earth, so equating these we get: * EQN:(1/375402)^2r=9.8(R/r)^2 so * EQN:r^3=9.8(R^2)(375402)^2 This gives an answer of 382517km, which is amazingly close to the figure quoted on WikiPedia of an average centre-centre distance of 384,403km. Accurate to 0.1%. Finally, Orbital Velocity is given by EQN:v=\omega{r} , and that now works out as 382517/375402 km/s, or almost exactly 1.02km/s. ---- Part of the Farrago of Fragments.